1.0 Refraction of Light
When a ray of light travels from one transparent medium to another, it undergoes a change in its direction at the interface of the two media. This phenomenon of the change in direction of the path of light as it passes from one optical medium to another is called Refraction.
The Laws of Refraction
- The incident ray, the refracted ray, and the normal at the point of incidence, all lie in the same plane.
- Snell's Law: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media.
Snell's Law & Refractive Index
$$_1\mu_2 = \frac{\sin i}{\sin r} = \frac{v_1}{v_2}$$
Where: $_1\mu_2$ is the refractive index of medium 2 w.r.t medium 1, $i$ = angle of incidence, $r$ = angle of refraction.
1.1 Effects of Speed on Refraction
The fundamental cause of refraction is the change in the speed of light when it enters a different medium:
- Rarer to Denser: Light slows down and bends towards the normal ($\angle i > \angle r$).
- Denser to Rarer: Light speeds up and bends away from the normal ($\angle i < \angle r$).
- Normal Incidence: If light enters at $90^\circ$ to the surface ($i = 0^\circ$), it passes undeviated ($r = 0^\circ$).
During refraction, the speed and wavelength of light change, but the frequency remains unchanged because it is a characteristic of the source of light. This is a very common one-mark question in ICSE boards!
The speed of light in air is $3 \times 10^8\,m/s$ and in glass it is $2 \times 10^8\,m/s$. Calculate the absolute refractive index of glass.
Solution:
1. Formula: $\mu = \frac{C}{V}$ (Speed in vacuum / Speed in medium).
2. Substitution: $\mu = \frac{3 \times 10^8}{2 \times 10^8}$.
3. Calculation: $\mu = 1.5$.
Final Answer: The refractive index of glass is 1.5. (Note: Refractive index has no units).
A swimming pool always appears shallower than it actually is because of refraction! Light rays coming from the bottom of the pool bend away from the normal as they exit the water, creating a "virtual bottom" for our eyes.
2.0 Refraction through a Rectangular Glass Slab
When light passes through a rectangular glass slab, it undergoes refraction at two parallel surfaces. A unique feature of this process is that the emergent ray is parallel to the incident ray but shifted slightly to the side. This shift is known as Lateral Displacement.
Lateral Displacement
The perpendicular distance between the path of the incident ray (extended forward) and the emergent ray is called Lateral Displacement.
Factors affecting Lateral Displacement:
- Thickness ($t$) of the slab: More thickness = More displacement.
- Angle of incidence ($i$): Larger angle = More displacement.
- Refractive Index ($\mu$): Higher refractive index = More displacement.
- Wavelength ($\lambda$): Displacement is more for violet light (short $\lambda$) than for red light (long $\lambda$).
2.1 Multiple Media and Principle of Reversibility
According to the Principle of Reversibility, if the path of a light ray is reversed after several refractions/reflections, it will exactly retrace its entire path. Mathematically:
$$_a\mu_g = \frac{1}{_g\mu_a}$$
Real and Apparent Depth
When looking from air into a denser medium (like water), the object appears closer:
$$\mu = \frac{\text{Real Depth}}{\text{Apparent Depth}}$$
Shift = Real Depth $-$ Apparent Depth $= \text{Real Depth} \times (1 - \frac{1}{\mu})$
The formula $\mu = \frac{\text{Real Depth}}{\text{Apparent Depth}}$ is valid only when the observation is made nearly normally (vertically downwards). If viewed at a large angle, the apparent depth changes and the math becomes more complex!
A postage stamp kept under a glass block of thickness $9\,cm$ appears to be raised by $3\,cm$. Calculate the refractive index of glass.
Solution:
1. Real Depth: $9\,cm$.
2. Shift: $3\,cm$.
3. Apparent Depth: Real Depth $-$ Shift $= 9 - 3 = 6\,cm$.
4. Refractive Index ($\mu$): $\frac{\text{Real Depth}}{\text{Apparent Depth}} = \frac{9}{6} = \mathbf{1.5}$.
Final Answer: The refractive index of glass is 1.5.
Because of atmospheric refraction, we see the Sun about 2 minutes before actual sunrise and 2 minutes after actual sunset. The Earth's atmosphere bends the sunlight towards us even when the Sun is slightly below the horizon!
3.0 Refraction through a Prism
A Prism is a transparent refracting medium bounded by five plane surfaces, with two triangular bases and three rectangular lateral surfaces. Unlike a glass slab where the surfaces are parallel, the refracting surfaces of a prism are inclined at an angle called the Angle of Prism ($A$). This causes the light to deviate towards the base of the prism.
Angle of Deviation ($\delta$)
The angle between the direction of the incident ray and the direction of the emergent ray is called the Angle of Deviation.
Factors affecting the Angle of Deviation:
- Angle of Incidence ($i$): As $i$ increases, $\delta$ first decreases, reaches a minimum, and then increases.
- Angle of Prism ($A$): Larger $A$ leads to a larger $\delta$.
- Refractive Index ($\mu$): Higher $\mu$ (material density) leads to a larger $\delta$.
- Color of Light: Deviation is maximum for Violet and minimum for Red light.
The Prism Equations
$$A = r_1 + r_2$$
$$\delta = (i_1 + i_2) - A$$
Where: $i_1$ = angle of incidence, $i_2$ = angle of emergence, $r_1$ and $r_2$ = internal angles of refraction.
3.1 Position of Minimum Deviation ($\delta_{min}$)
In the position of Minimum Deviation, the light ray passes symmetrically through the prism. This special case is highly important for ICSE numericals:
- The angle of incidence is equal to the angle of emergence ($i_1 = i_2 = i$).
- The angle of refraction at the first face is equal to the angle of refraction at the second face ($r_1 = r_2 = r = A/2$).
- The refracted ray inside the prism is parallel to the base (for an equilateral prism).
The $\delta-i$ graph is a parabolic curve, not a straight line. Always remember that for every value of $\delta$ (except $\delta_{min}$), there are two angles of incidence ($i_1$ and $i_2$) that produce the same deviation.
A ray of light is incident at an angle of $48^\circ$ on the face of an equilateral prism. It suffers minimum deviation. Calculate the angle of minimum deviation.
Solution:
1. Equilateral Prism: Angle of Prism ($A$) = $60^\circ$.
2. At Minimum Deviation: $i_1 = i_2 = 48^\circ$.
3. Formula: $\delta_{min} = (i_1 + i_2) - A$.
4. Calculation: $\delta_{min} = (48 + 48) - 60 = 96 - 60 = \mathbf{36^\circ}$.
Final Answer: The angle of minimum deviation is $36^\circ$.
Sir Isaac Newton was the first to prove that a prism doesn't "color" the light, but rather spreads out the white light into its constituent colors (Spectrum). He proved this by placing a second inverted prism which recombined the colors back into white light!
4.0 Critical Angle and Total Internal Reflection (TIR)
When light travels from an optically denser medium to an optically rarer medium, it bends away from the normal. As the angle of incidence increases, the angle of refraction also increases until it reaches a specific point where light no longer exits the medium but reflects back entirely.
Critical Angle ($C$)
The Critical Angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is exactly $90^\circ$.
Factors affecting Critical Angle:
- Color of Light: Critical angle is maximum for Red and minimum for Violet light.
- Temperature: As temperature increases, the refractive index decreases, and the critical angle increases.
4.1 Total Internal Reflection (TIR)
If the angle of incidence in the denser medium is increased beyond the critical angle ($i > C$), the ray does not pass into the rarer medium but is strictly reflected back into the denser medium following the laws of reflection. This is Total Internal Reflection.
For TIR to occur, two conditions must be met simultaneously:
1. Light must travel from a Denser medium to a Rarer medium.
2. The angle of incidence must be greater than the critical angle ($i > C$).
Relationship: $\mu$ and Critical Angle
$$\mu = \frac{1}{\sin C}$$
Where: $\mu$ is the refractive index of the denser medium w.r.t the rarer medium.
The critical angle for a certain transparent medium is $45^\circ$. Calculate the refractive index of the medium.
Solution:
1. Given: Critical Angle ($C$) = $45^\circ$.
2. Formula: $\mu = 1 / \sin C$.
3. Calculation: $\mu = 1 / \sin 45^\circ = 1 / (1/\sqrt{2}) = \sqrt{2} \approx \mathbf{1.414}$.
Final Answer: The refractive index is 1.414.
Optical fibers, which carry the high-speed internet to your home, work entirely on the principle of Total Internal Reflection! Light signals bounce inside the thin glass fibers thousands of times without losing energy, allowing data to travel across oceans in milliseconds.
5.0 Total Reflecting Prisms
A Total Reflecting Prism is a right-angled isosceles prism (having angles $45^\circ$, $90^\circ$, and $45^\circ$). These prisms are often used to replace plane mirrors in optical instruments because they are much more efficient and produce sharper images without the "ghosting" effects of silvered glass.
Why Use Prisms over Mirrors?
- 100% Reflection: In TIR, the entire light energy is reflected, whereas a mirror absorbs some light.
- Durability: The reflecting surface of a prism is internal and cannot be scratched or tarnished.
- No Multiple Images: Mirrors often produce a faint "front-surface" reflection and a main "back-surface" reflection; prisms produce only one clear image.
5.1 Three Main Actions of a $45^\circ-90^\circ-45^\circ$ Prism
Depending on which face the light is incident upon, the prism can manipulate the ray in three ways:
- To Deviate a Ray through $90^\circ$: Light enters normally through one cathetus face and strikes the hypotenuse at $45^\circ$. Since $45^\circ > C$ (for glass, $C \approx 42^\circ$), TIR occurs. Used in Periscopes.
- To Deviate a Ray through $180^\circ$: Light enters normally through the hypotenuse face. It suffers TIR twice inside the prism. Used in Binoculars.
- To Erect an Inverted Image without Deviation: Light enters at an angle through one face and is reflected by the hypotenuse. Used in Slide Projectors.
The $60^\circ$ Equilateral Prism Case
For an equilateral prism ($A=60^\circ$):
If $i = 0^\circ$ at one face, the ray hits the second face at $60^\circ$.
Since $60^\circ > 42^\circ$, the ray will undergo TIR at the second face instead of emerging.
When drawing ray diagrams, if a ray strikes a surface normally ($\angle i = 0^\circ$), it must be drawn straight without any bending. Refraction only happens when it reaches the next surface at an angle.
A ray of light is incident normally on one face of a $45^\circ-90^\circ-45^\circ$ glass prism. Why does it suffer Total Internal Reflection at the hypotenuse? (Given $_a\mu_g = 1.5$)
Solution:
1. Calculate Critical Angle ($C$): $\sin C = 1/\mu = 1/1.5 = 0.666$. Therefore, $C \approx 42^\circ$.
2. Determine Angle of Incidence ($i$): Since the ray is normal to one side, it strikes the hypotenuse at an angle of $45^\circ$.
3. Condition Check: Since $i (45^\circ) > C (42^\circ)$, the light cannot pass into the air.
Final Answer: The light undergoes TIR because the angle of incidence exceeds the critical angle.
A diamond's brilliant sparkle is due to its extremely low critical angle ($\approx 24.4^\circ$). Because the critical angle is so small, light entering the diamond is much more likely to suffer multiple Total Internal Reflections before finally exiting, making the diamond glow from within!
6.0 Atmospheric Refraction and Natural Phenomena
The Earth's atmosphere is not a uniform medium. Its density and temperature vary with altitude, which in turn causes the refractive index of air to change. As light from celestial bodies enters the atmosphere, it passes through layers of different optical densities, leading to Atmospheric Refraction.
Key Atmospheric Effects
- Apparent Position of Stars: Stars appear slightly higher than their actual position because light bends towards the normal as it enters the increasingly denser lower atmosphere.
- Twinkling of Stars: Due to the continuous movement of air and changing temperature, the refractive index of the atmosphere fluctuates. This causes the apparent position and intensity of starlight to shift rapidly.
- Early Sunrise & Delayed Sunset: The Sun is visible for about 2 minutes before actual sunrise and 2 minutes after sunset due to the bending of light around the curvature of the Earth.
6.1 The Mirage
A Mirage is an optical illusion observed in hot deserts or over coal-tarred roads on a hot summer day. It is caused by the Total Internal Reflection of light in the layers of air near the ground.
On a hot day, the air near the ground becomes much hotter and rarer than the air above it. A ray of light from a tall object (like a tree) traveling downwards passes from denser to rarer layers. At a certain layer, the angle of incidence exceeds the Critical Angle, causing TIR. To the observer, the light appears to come from the ground, creating an inverted image that looks like a reflection in a pool of water.
This is a favorite ICSE reasoning question! Stars are point sources of light, so even a slight shift is noticeable. Planets are extended sources (closer to Earth), so the total variations in light from different points on the planet's disc average out to zero, resulting in a steady glow.
Explain why the duration of daylight on Earth is approximately 4 minutes longer than it would be without an atmosphere.
Solution:
1. When the Sun is slightly below the horizon, the sunlight enters the Earth's atmosphere (denser medium) from space (rarer medium).
2. The light rays undergo refraction and bend towards the normal.
3. This causes the Sun to appear above the horizon for 2 minutes before it actually rises and 2 minutes after it has actually set.
Final Answer: Total extra daylight = 2 mins (sunrise) + 2 mins (sunset) = 4 minutes.
There is a cold-weather version of a mirage called Looming! In polar regions, the air near the ground is much colder and denser than the air above. This causes light to bend downwards, making ships or icebergs appear to be floating high in the sky!